# Unital commutative ring and distinct maximal ideals.

Let $R$ be a unital commutative ring and $m_1,m_2$ distinct maximal ideals. Prove that $$\frac{R}{m_1m_2}\simeq\frac{R}{m_1} \times \frac{R}{m_2}.$$

I think something like this homomorphism might work: $\phi(r)=(r+m_1,r+m_2)$.

Now $\ker\phi=\cdots=m_1\cap m_2$, but it's not clear if it is a surjection.

Also it's true that

If $m_1,m_2$ are maximal ideals of $R$, then $m_1m_2=m_1\cap m_2$.

• Chinese remainder theorem in generalized form. – egreg Dec 28 '13 at 20:56
• Note the statement here en.wikipedia.org/wiki/… that if the ideals $m_i$ are pairwise coprime (meaning $m_ + m_j = R$ for $i \ne j$, which holds if they are maximal) then their produtc coincides with their intersection. – Andreas Caranti Dec 28 '13 at 22:17

The map $\phi : R \to \frac{R}{m_1}\times \frac{R}{m_2}$, $\phi(r) = (r + m_1, r + m_2)$ is the correct homomorphism to consider.

As $m_1$ and $m_2$ are comaximal, $m_1 + m_2 = R$. As $R$ is unital $1 \in R$. So there are $a\in m_1$, $b \in m_2$ such that $a + b = 1$.

Exercise: Show that $\phi(a) = (0 + m_1, 1 + m_2)$ and $\phi(b) = (1 + m_1, 0 + m_2)$.

We have $\phi(a) = (a + m_1, a + m_2) = (0 + m_1, 1 - b + m_2) = (0 + m_1, 1 + m_2)$. Likewise for $\phi(b)$.

Now, any $(r_1 + m_1, r_2 + m_2) \in \frac{R}{m_1}\times\frac{R}{m_2}$ can be written as a linear combination of $\phi(a)$ and $\phi(b)$.

Exercise Show that there is $x \in R$ such that $\phi(x) = (r_1 + m_1, r_2 + m_2)$.

We have $(r_1 + m_1, r_2 + m_2) = (r_1 + m_1, 0 + m_2) + (0 + m_1, r_2 + m_2) = r_1\phi(a) + r_2\phi(b)$ so if we set $x = r_1a+r_2b$, we see that $\phi(x) = (r_1 + m_1, r_2 + m_2)$.

So the map $\phi$ is surjective, so by the first isomorphism theorem

$$\frac{R}{\ker\phi} \cong \frac{R}{m_1}\times\frac{R}{m_2}.$$

As you've already determined, $\ker\phi = m_1\cap m_2$. As $m_2$ is an ideal, $Rm_2 \subseteq m_2$, so $m_1m_2 \subseteq m_2$. Likewise, $m_1m_2 \subseteq m_1$ so $m_1m_2 \subseteq m_1\cap m_2$.

Exercise: Show $m_1\cap m_2 \subseteq m_1m_2$. Hint, use the fact that we have $a \in m_1$ and $b \in m_2$ such that $a + b = 1$, then pick $r \in m_1\cap m_2$.

As $a + b = 1$, $ra + rb = r$. As $r \in m_1\cap m_2 \subseteq m_2$ and $a \in m_1$, $ra \in m_1m_2$. Likewise, $rb \in m_1m_2$, so $r = ra + rb \in m_1m_2$.

Then $m_1\cap m_2 = m_1m_2$ so we finally obtain $$\frac{R}{m_1m_2} \cong \frac{R}{m_1}\times\frac{R}{m_2}.$$

• If you are stuck on an exercise, you can put your mouse over the grey box beneath it to reveal the solution. I recommend you try the exercises yourself first. – Michael Albanese Dec 28 '13 at 23:30

The fact that $m_1\cap m_2=m_1m_2$ when $m_1\ne m_2$ is really easy to prove. Now the morphism $$R\to \frac{R}{m_1}\times\frac{R}{m_2}$$ defined by $f(r)=(r+m_1,r+m_2)$ is surjective because $m_1+m_2=R$, so you can write $1=x+y$, with $x\in m_1$ and $y\in m_2$.

If $a,b\in R$, we need to find $r\in R$ such that $r-a\in m_1$ and $r-b\in m_2$. But $$a-b=(a-b)1=(a-b)x+(a-b)y$$ so $$r=a-(a-b)x=b+(a-b)y$$ is the element we're looking for.

The kernel of the above morphism is obviously $m_1\cap m_2=m_1m_2$.

Nothing really different from the well known Chinese Remainder Theorem.